Math Problem Statement

The foci of a hyperbola coincide with the foci of the ellipse x^2/100+y^2/36=1 Find the equation of the hyperbola, if its eccentricity is 4.

Solution

To find the equation of the hyperbola, we proceed as follows:

Step 1: Foci of the ellipse

The given ellipse equation is:

x2100+y236=1\frac{x^2}{100} + \frac{y^2}{36} = 1

This is an ellipse centered at the origin (0,0)(0, 0) with the following properties:

  • Semi-major axis: a=100=10a = \sqrt{100} = 10
  • Semi-minor axis: b=36=6b = \sqrt{36} = 6

The distance from the center to each focus is given by:

c=a2b2=10262=10036=64=8c = \sqrt{a^2 - b^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8

Thus, the foci of the ellipse are located at (±8,0)(\pm 8, 0).

Step 2: Hyperbola shares the same foci

For a hyperbola, the relationship between aa, bb, and cc is given by:

c2=a2+b2c^2 = a^2 + b^2

The eccentricity of the hyperbola is given as e=4e = 4. The formula for eccentricity is:

e=cae = \frac{c}{a}

Rearranging to find aa, we get:

a=cea = \frac{c}{e}

Substituting c=8c = 8 and e=4e = 4:

a=84=2a = \frac{8}{4} = 2

Step 3: Solve for b2b^2

Using the hyperbola relationship c2=a2+b2c^2 = a^2 + b^2, substitute c=8c = 8 and a=2a = 2:

82=22+b28^2 = 2^2 + b^2

64=4+b264 = 4 + b^2

b2=644=60b^2 = 64 - 4 = 60

Step 4: Write the hyperbola equation

Since the foci lie on the xx-axis, the hyperbola opens horizontally. The standard equation of a horizontally opening hyperbola is:

x2a2y2b2=1\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1

Substituting a2=4a^2 = 4 and b2=60b^2 = 60:

x24y260=1\frac{x^2}{4} - \frac{y^2}{60} = 1

Final Answer:

The equation of the hyperbola is:

x24y260=1\frac{x^2}{4} - \frac{y^2}{60} = 1


Would you like a detailed derivation or additional explanations?
Here are 5 related questions to explore further:

  1. What happens to the shape of a hyperbola if its eccentricity increases further?
  2. How is the equation of a hyperbola derived from its geometric definition?
  3. Can you find the equation of the conjugate hyperbola of the given hyperbola?
  4. What are the asymptotes of the hyperbola x24y260=1\frac{x^2}{4} - \frac{y^2}{60} = 1?
  5. How do the properties of ellipses and hyperbolas differ geometrically and algebraically?

Tip: Always verify eccentricity and the foci of conic sections to ensure correctness when switching between ellipses and hyperbolas.

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Math Problem Analysis

Mathematical Concepts

Hyperbolas
Ellipses
Eccentricity
Conic Sections

Formulas

Equation of ellipse: x^2/a^2 + y^2/b^2 = 1
Distance to foci (ellipse): c = √(a^2 - b^2)
Equation of hyperbola: x^2/a^2 - y^2/b^2 = 1
Eccentricity of hyperbola: e = c/a

Theorems

Relationship between foci, semi-major axis, and semi-minor axis for ellipses
Relationship between eccentricity and foci in hyperbolas

Suitable Grade Level

Grades 10-12